#question
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7⅞
?
multiply maybe?
multiply what sir
equations
all
Why thou
try eq1eq2, eq2eq3 and so on
Maybe you'll get the answer?
How do you even know whats right?
We dont know the asnwer
U just gonna have to roll the dice on this one pal
You might win
no I mean like
what is multiply all equation
ax + by = 1
bx + cy = 1
cx + ay = 1
Proof:
ab + bc + ca = a^2 + b^2 + c^2
multiply eqn two at a time or 3 at a time
so like
(ax+by)(bx+cy)
ya
And with diff combination until I get ab + bc + ca = a^2……
you try
why
It is add addition
And = 1
If I swap side after I simplify them
It’s gonna be negative
And division
So even if I found the combination , it will be …
ab + bc + ca = a^2 - b^2 + c^2
or ab + bc + ca = 1/a^2…….
So, you would use the method of elimination and substitution.
[
x = \frac{1 - by}{a}
]
[
b\left(\frac{1 - by}{a}\right) + cy = 1
]
[
b(1 - by) + acy = a
]
[
b - b^2y + acy = a
]
[
(b^2 - ac)y = b - a
]
[
y = \frac{b - a}{b^2 - ac}
]
asura Ψ
That's a start I believe.
Prob
And let me try sub it into an equation
ax+by=bx+cy=cx+ay
ax+by=bx+cy and bx+cy=cx+ay
ax-bx=cy-by and bx-cx=ay-cy
(a-b)x=(c-b)y and (b-c)x=(a-c)y
x/y=(c-b)/(a-b) and y/x=(b-c)/(a-c)
x=c-b and y=a-b
x=a-c and y=b-c
c-b=a-c implies 2c=a+b
a-b=b-c implies 2b=a+c
2c-2b=a-a+b-c
2c-2b=b-c
2c+c=b+2b
3c=3b
c=b